Problem: Simplify and expand the following expression: $ \dfrac{1}{r + 6}- \dfrac{1}{3r - 6}- \dfrac{3r}{r^2 + 4r - 12} $
Answer: First find a common denominator by finding the least common multiple of the denominators. Try factoring the denominators. We can factor a $3$ out of denominator in the second term: $ \dfrac{1}{3r - 6} = \dfrac{1}{3(r - 2)}$ We can factor the quadratic in the third term: $ \dfrac{3r}{r^2 + 4r - 12} = \dfrac{3r}{(r + 6)(r - 2)}$ Now we have: $ \dfrac{1}{r + 6}- \dfrac{1}{3(r - 2)}- \dfrac{3r}{(r + 6)(r - 2)} $ The least common multiple of the denominators is: $ (r + 6)(r - 2)$ In order to get the first term over $(r + 6)(r - 2)$ , multiply by $\dfrac{3(r - 2)}{3(r - 2)}$ $ \dfrac{1}{r + 6} \times \dfrac{3(r - 2)}{3(r - 2)} = \dfrac{3(r - 2)}{(r + 6)(r - 2)} $ In order to get the second term over $(r + 6)(r - 2)$ , multiply by $\dfrac{r + 6}{r + 6}$ $ \dfrac{1}{3(r - 2)} \times \dfrac{r + 6}{r + 6} = \dfrac{r + 6}{(r + 6)(r - 2)} $ In order to get the third term over $(r + 6)(r - 2)$ , multiply by $\dfrac{3}{3}$ $ \dfrac{3r}{(r + 6)(r - 2)} \times \dfrac{3}{3} = \dfrac{9r}{(r + 6)(r - 2)} $ Now we have: $ \dfrac{3(r - 2)}{(r + 6)(r - 2)} - \dfrac{r + 6}{(r + 6)(r - 2)} - \dfrac{9r}{(r + 6)(r - 2)} $ $ = \dfrac{ 3(r - 2) - (r + 6) - 9r} {(r + 6)(r - 2)} $ Expand: $ = \dfrac{3r - 6 - r - 6 - 9r}{3r^2 + 12r - 36} $ $ = \dfrac{-7r - 12}{3r^2 + 12r - 36}$